Mathematical relationship between two line segments perpendicular to a circle

Hello,

What is the mathematical relationship between the length of two line segments, each of which extends radially (perpendicularly) from a separate point on a circle to a line which is tangent to a third point on the circle?

For example, the following diagram shows a circle with origin o and radius r (not = 1) which is centered on the intersection of horizontal axis x and vertical axis y. Line z is tangent to the circle at point p which is the top intersection of the circle and vertical axis y. Line segments s & t (red & green respectively) each begin on the circle and extend radially to line x. Angle a is formed by the vertical axis y and the radial extension of line segment s while angle b is formed by the radial extensions of line segments s & t. Arc d extends from the origin of line segment s to the origin of line segment t.


Questions:

1) What is the mathematical relationship between the length of line segments s & t?

2) What is the length of arc d (may be expressed in radians)?

Thanks
Steve_BradyAsked:
Who is Participating?
I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

ozoCommented:
1)
 s=(secant(a)-1)*r  
t = (secant(a+b)-1)*r
2)
d = r*b
0
JohnBusiness Consultant (Owner)Commented:
If it helps you to think about this from first principles, you have two right angle triangles Y, o, s and Y, o, t with a third triangle embedded s, t and a segment of the tangent. Since two of the triangles are right angled, you can work out all the numbers from first principles if you wish. It comes out to #1 in the above post.

.... Thinkpads_User
0
Steve_BradyAuthor Commented:
Thanks for the response.

Could you show the derivation please?

I understand that

        secant = hypotenuse/adjacent

but I have not been able to determine how you arrived at that solution.
0
Exploring SharePoint 2016

Explore SharePoint 2016, the web-based, collaborative platform that integrates with Microsoft Office to provide intranets, secure document management, and collaboration so you can develop your online and offline capabilities.

ozoCommented:
adjacent = r
hypotenuse = r+s or r+t
0
phoffricCommented:
>> 1. What is the mathematical relationship between the length of line segments s & t?
To answer this equation, you should express s and t in terms of each other.

Using ozo's equations:
s=(secant(a)-1)*r  
t = (secant(a+b)-1)*r

Easiest way to express one in terms of the other is to divide lhs and rhs of the two equations:

s/t = { (secant(a)-1)*r } / { (secant(a+b)-1)*r }

You can cancel the r factor in the numerator and denominator leaving:

s/t = (secant(a)-1) / (secant(a+b)-1)

This means that for a constant pair of angles, a, b, the ratio of s/t is unaffected by the size of the circle since the r factor has been removed.

You can now write:
s = t * (secant(a)-1) / (secant(a+b)-1)


====================

Just to make sure that you understand how to derive ozo's equations:

secant (angle) = hypotenuse/adjacent

adjacent in both right triangles is just the vertical line, OP, which is just the radius of the circle. So, OP = r.

hyp associated with angle a is r+s
hyp associated with angle (a+b) is r+t

sec(a)     = (r+s)/r ==> (r+s) = r sec(a)     ==> s = r sec(a) - r = r( sec(a) - 1)
sec(a+b) = (r+t)/r ==> (r+t) = r sec(a+b) ==> t = r sec(a+b) - r = r( sec(a+b) - 1)
0

Experts Exchange Solution brought to you by

Your issues matter to us.

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Start your 7-day free trial
Steve_BradyAuthor Commented:
Many thanks.
0
It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
Math / Science

From novice to tech pro — start learning today.